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Electric Charges And Fields

Question
CBSEENPH12039211

(a) Deduce the expression for the torque acting on a dipole of dipole moment p in the presence of a uniform electric field E⃗.

(b) Consider two hollow concentric spheres, S1 and S2, enclosing charges 2Q and 4Q respectively as shown in the figure.

(i) Find out the ratio of the electric flux through them.

(ii) How will the electric flux through the sphere S1 change if a medium of dielectric constant 'εr' is introduced in the space inside S1 in place of air ?

 Deduce the necessary expression.  


Solution

a) Dipole in a uniform electric field:

Consider an electric dipole consisting of charges −q and +q and of length 2a placed in a uniform electric field making an angle θ with electric field.

Forces acting on the two charges of the dipole, are +qE and -qE.

That is, the net force on the dipole is equal and opposite.

So, Force = 0

Two forces are equivalent to torque having magnitude given by,
straight tau space equals space left parenthesis qE right parenthesis AC

space space equals space straight q space straight E. space 2 space straight a space sin space straight theta

space space equals space straight p space straight E space sin space straight theta 
Therefore, torque acting on the dipole is given by, 
straight tau with rightwards harpoon with barb upwards on top space equals space p with rightwards harpoon with barb upwards on top space x space E with rightwards harpoon with barb upwards on top

b) i)  Charge enclosed by sphere S1 = 2Q

Charge enclosed by sphere S2 = 2Q + 4Q = 6Q

Now, using Gauss law, electric flux enclosed by sphere S1 and S2 is given by, 
italic space italic space italic space italic space italic space italic space ϕ subscript 1 equals fraction numerator 2 Q over denominator epsilon subscript 0 end fraction a n d space ϕ subscript 2 equals fraction numerator 6 Q over denominator epsilon subscript 0 end fraction

Ratio space of space electric space flux space is comma space

space space space space space space space space space space ϕ subscript 1 over ϕ subscript 2 equals space fraction numerator fraction numerator 2 Q over denominator epsilon subscript 0 end fraction over denominator fraction numerator 6 Q over denominator epsilon subscript 0 end fraction end fraction space equals space 1 third 

ii) If a medium of dielectric constant 'εr' is introduced in the space inside S1 in place of air, electric flux becomes
ϕ subscript 1 apostrophe equals fraction numerator 2 Q over denominator epsilon subscript r end fraction
italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space epsilon subscript r italic space end subscript italic greater than italic space epsilon subscript o italic space
T h e r e f o r e italic comma italic space

italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space ϕ subscript italic 1 italic apostrophe italic space italic less than italic space ϕ subscript italic 1

 
That is, electric flux decreases.